Q.A year is selected at random. What is the probability that

Step 1. Per-type probabilities (from the classical reasoning of a non-leap/leap year's extra days). $P(53\text{ Sun}/\text{non-leap})=\dfrac17$, $P(53\text{ Sun}/\text{leap})=\dfrac27$.

Step 2. Weight by how common each type of year is. Using the standard simplified convention $P(\text{leap})=\dfrac14$, $P(\text{non-leap})=\dfrac34$ (one leap year in every four).

Step 3. Part (i) -- total probability of 53 Sundays. By Total Probability: $P(53\text{ Sun})=P(\text{non-leap})P(53\text{ Sun}/\text{non-leap})+P(\text{leap})P(53\text{ Sun}/\text{leap})=\dfrac34\times\dfrac17+\dfrac14\times\dfrac27=\dfrac{3}{28}+\dfrac{2}{28}=\dfrac{5}{28}$. …

Total Probability weighting the two year-types by their frequency, using the Multiplication The …

• Answering part (ii) with the conditional probability $P(53\text{ Sun}/\text{leap})=\tfrac27$ instead of the JOINT probability $P(\text{leap AND }53 …